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Math Help - Homogeneous DE of the first order

  1. #1
    MHF Contributor arbolis's Avatar
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    Homogeneous DE of the first order

    I must solve \dot x + x\cos t=0.
    I realize it's separable, and solving it I reach x=e^{-\sin (t) + C} where C is a constant.

    However, solving it via the integrating factor method, I do not reach the same result.
    The IF is e^{\sin t}.
    Multiplying both sides by it, I get \frac{dx}{dt} e^{\sin t}+x\cos (t) e^{\sin t}=0.
    Integrating with respect to t, I get e^{\sin t}x=0.
    Now clearly x\equiv 0.
    What am I doing wrong?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I must solve \dot x + x\cos t=0.
    I realize it's separable, and solving it I reach x=e^{-\sin (t) + C} where C is a constant.

    However, solving it via the integrating factor method, I do not reach the same result.
    The IF is e^{\sin t}.
    Multiplying both sides by it, I get \frac{dx}{dt} e^{\sin t}+x\cos (t) e^{\sin t}=0.
    Integrating with respect to t, I get e^{\sin t}x=0.
    Now clearly x\equiv 0.
    What am I doing wrong?
    What I think we need to do at \frac{\,d}{\,dt}\left[xe^{\sin t}\right]=0 is integrate both sides and say that xe^{\sin t}={\color{red}C} in order to avoid getting the trivial solution for an answer. From here then, we get x=Ce^{-\sin t} which is the answer you got from solving it using separation of variables.

    Does this help?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    What I think we need to do at \frac{\,d}{\,dt}\left[xe^{\sin t}\right]=0 is integrate both sides and say that xe^{\sin t}={\color{red}C} in order to avoid getting the trivial solution for an answer. From here then, we get x=Ce^{-\sin t} which is the answer you got from solving it using separation of variables.

    Does this help?
    Bright!
    Yes it helps, a lot.
    By the way your tutorial is really great! I'm going to use it often these days.
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